Red and Blue

Initially :

${\color{#D61F06}\text{Red Pens = }4x} \quad and \quad {\color{#3D99F6}\text{Blue pens = }3x}$

Finally :

$\dfrac{4x - 1}{3x - 2} = \dfrac{3}{2}$

$\implies 8x - 2 = 9x - 6$

$\implies x = 4$

$\therefore {\color{#D61F06}\text{Red Pens}} = 4x = 4(4) = \color{#D61F06}16$

Let the number of red and blue pens in the box be $4x$ and $3x$ respectively. The number of red and blue pens in the hypothetical scenario shall be $3y$ and $2y$ respectively so that $4x-1=3y$ and $3x-2=2y$ . Then $y=\dfrac{3x-2}{2}$ .

$4x-1=3\left(\dfrac{3x-2}{2}\right)$

$2(4x-1)=3(3x-2)$

$8x-2=9x-6$

$x=4$

Thus there are $4\times 4=\boxed{16}$ red pens in the box.